The distance at which the man slips is 0.3 m
Newton's Second Law, F = ma, is used to calculate the braking distance. By dividing the mass of the car by the gravitational acceleration, one may determine its weight. The weight of the car multiplied by the coefficient of friction equals the brake force.
Given-
mass of man= 70 kg
frictional coefficient μ=0.02
mass of body thrown= m2 = 3kg
let s be the stopping distance
we know that frictional force = F= μN
=μMg= 0.02 x 70 x 10
=14 N
∴acceleration, a= 14/70 = 0.2 m/s²
now on applying conservation of linear momentum
pi=pf pi=0 (initially at rest)
0=m1v1-m2v2 (v1= velocity of man) (v2=velocity of body= 8m/s
v1= m2v2 /m1= 0.3 m/s
we know,
v²- u² = -2as
0- (0.3) ²= -2 x 0.2 x 5
s= 0.09/0.4 ≈ 0.3 m
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Answer: v = 4.4 m/s
Explanation:
In the absence of friction, the total mechanical energy will be constant
KE₀ + PE₀ = KE₁ + PE₁
0 + mg(6) = ½mv₁² + mg(5)
½mv₁² = mg(6 - 5)
v = √(2g(1)) = 4.4 m/s
Answer:
a) yield strength
b) modulus of elasticity
strain calculation
strain for offset yield point
=0.0046-0.002 = 0.0026
now, modulus of elasticity
= 184615.28 MPa = 184.615 GPa
c) tensile strength
d) percentage elongation
e) percentage of area reduction
